EE215: Fundamentals of Electrical Engineering
Midterm Exam 1
January 26, 2004
Time: in class, 50 minutes
Class: _
EE215
_____________
Section:
____________________
Name: ____________________
Student ID:
____________________
Please show all steps in solving the problems. Clearly mark your final answers and don’t forget
the units.
Problem 1 (30 points)
(a) Determine
v
o
in the circuit above!
(20 points)
Answer:
This is similar to problem 3.17.
R
eq
= 24k
Ω
+ (60k
Ω
 (30k
Ω
+ 60k
Ω
)) = 24k
Ω
+ (60k
Ω
 90k
Ω
) = 24k
Ω
+ 36k
Ω
= 60k
Ω
.
It follows that the voltage between
a
and
d
is 36/60 · 100V = 60V,
and thus
v
o
= 60/90 · 60V =
40V
.
(b) What is the total power dissipated in this circuit?
(10 points)
Answer:
The total power dissipated is equal to the total power developed in the voltage source,
P
=
v
2
/
R
eq
= 10000/60000 W =
1/6 W
.
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Problem 2 (35 points)
Find the voltage
v
y
in the circuit above!
Answer:
This is a mirrored and simplified version of Problem 2.25.
Let’s call the current through the 200
Ω
resistor
i
200
.
(1)
29
i
x
+
i
x
=
i
200
(KCL)
(2)
25 – 500 · 29
i
x
–
v
y
– 200 ·
i
200
= 0
(KVL)
(3)
16 – 10,000 ·
i
x
– 200 ·
i
200
= 0
(KVL)
From (1) we obtain
i
200
= 30
i
x
(2’)
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 Spring '05
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 Electrical Engineering, Power, Resistor, SEPTA Regional Rail, Electrical resistance, Series and parallel circuits

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